ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS

ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for all p, or equivalently, with roots mod n for all n. It is known that f(x) cannot be irreducible but can be a product of two or more irreducible polynomials, and that if f(x) is a product of m > 1 irreducible polynomials, then its Galois group must be m-coverable, i.e. a union of conjugates of m proper subgroups, whose total intersection is trivial. We are thus led to a variant of the inverse Galois problem: given an m-coverable finite group G, find a Galois realization of G over the rationals Q by a polynomial f(x) Z[x] which is a product of m nonlinear irreducible factors (in Q[x]) such that f(x) has a root in Q p for all p. The minimal value m = 2 is of special interest. It is known that the symmetric group S n is 2-coverable if and only if 3 n 6, and the alternating group A n is 2-coverable if and only if 4 n 8. In this paper we solve the above variant of the inverse Galois problem for the 2-coverable symmetric and alternating groups, and exhibit an explicit polynomial for each group, with the help of the software packages MAGMA, PARI and GAP. 1. Introduction Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for all p, or equivalently, with roots mod n for all n. It is known that f(x) cannot be irreducible but can be a product of two or more irreducible polynomials, and that if f(x) is a product of m > 1 irreducible polynomials, then its Galois group must be m-coverable, i.e. a union of conjugates of m proper subgroups, whose total intersection is trivial. We are thus led to a variant of the inverse Galois problem: given an m-coverable finite group G, find a Galois realization of G over the rationals Q by a polynomial f(x) Z[x] which is a product of m nonlinear irreducible factors (in Q[x]) such that f(x) has a root in Q p for all p. The existence of such a polynomial can be established by means of the following proposition. Proposition 1.1. [6] Let K/Q be a finite Galois extension with Galois group G. The following are equivalent: (1) K is a splitting field of a product f(x) = g 1 (x) g m (x) of m irreducible polynomials of degree greater than one in Q[x] and f(x) has a root in Q p for all primes p. Date: October 21, 2010. 2000 Mathematics Subject Classification. Primary 11R32, Secondary 11Y40; 11R09. Key words and phrases. Galois group, symmetric group, alternating group, polynomial, roots modulo n. 1

2 DANIEL RABAYEV AND JACK SONN (2) G is the union of conjugates of m proper subgroups A 1,..., A m, the intersection of all these conjugates is trivial, and for all primes p of K, the decomposition group G(p) is contained in a conjugate of some A i. Remark. Condition (2) is evidently satisfied if G is m-coverable and all its decomposition groups are cyclic. This last condition holds automatically at all unramified primes. This remark is used in [6] to prove existence for any m-coverable finite solvable group. The minimal value m = 2 is of natural interest. All Frobenius groups are 2-coverable. In [6] it is proved that Galois realizations satisfying condition (2) of Prop. 1.1 exist for all nonsolvable Frobenius groups with m = 2. In this paper we focus on 2-coverable symmetric and alternating groups, which have been determined in [3] and [4]: S n is 2-coverable if and only if 3 n 6, and A n is 2-coverable if and only if 4 n 8. Polynomials for S 3 are exhibited in [1]. For S 4 and A 4 existence follows from the general theorem for solvable groups in [6]. For S 5, existence follows from a known quintic (see e.g. [5]). Until now, not even existence was known for S 6. In the present paper, with the help of the software packages MAGMA, GAP and PARI, we both prove existence and produce an explicit polynomial for each S n (3 n 6) and each A n (4 n 8): Theorem 1.2. Let G be a 2-coverable group which is either a symmetric group S n (i.e. 3 n 6), or an alternating group A n (i.e. 4 n 8). Then G is realizable over Q as the Galois group of a polynomial f(x) which is the product of two (nonlinear) irreducible polynomials in Q[x], such that f(x) has a root in Q p for all primes p. In what follows, we take a group G {S 4, S 5, S 6, A 4,..., A 8 }, and a 2-covering of G from [3] or [4], given by two subgroups U 1, U 2 of G. We then produce a polynomial f(x) Q[x] with Galois group G over Q which satisfies the condition that for every prime p of the splitting field K of f(x) over Q, the decomposition group G(p) is contained in a conjugate of U 1 or U 2. We then find two irreducible polynomials g 1 (x), g 2 (x) Q[x] whose splitting fields are contained in K, such that for each i = 1, 2, if η i K is a root of g i (x), then the Galois group of K over Q(η i ) is U i. It then follows that h(x) = g 1 (x)g 2 (x) has the desired property. The software packages MAGMA, GAP and PARI are used in the searches for f(x) and in the computation of g 1 (x), g 2 (x). 2. The symmetric groups S 4 : The 2-covering of S 4 in [3] is given by U 1 := the stabilizer in S 4 of one point and U 2 := the 2-Sylow subgroup of S 4. U 1 contains permutations of type 1 [2] and of type [3], and U 2 contains permutations of types [4] and [2, 2]. Since two elements of S n are conjugate if and only if they are of the same type, U 1, U 2 is a 2-covering of S 4. 1 If a 1,..., a k are natural numbers, and σ is a permutation, then σ is of type [a 1,..., a k ] if and only if σ is the product of k disjoint cycles of length a 1,..., a k respectively. In particular, a permutation of type [n] is an n-cycle.

2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS 3 Consider the polynomial f(x) = x 4 5x 2 + x + 4 with prime discriminant p = 2777. f(x) is irreducible mod 3, factors into the product (x + 4)(x 3 + 7x 2 + 1) of irreducible factors mod 11 and factors into the product (x + 8)(x + 19)(x 2 +19x+20) mod 23. These factorizations imply that the Galois group of f(x) contains a 4-cycle, 3-cycle and a transposition. This implies that the Galois group of f(x) (over Q), which is contained in S 4, is indeed S 4. Let K be the splitting field of f(x) over Q. We will show that for all primes p of K, the decomposition group G(p) is cyclic, hence contained in a conjugate of U 1 or U 2. The decomposition group of an unramified prime is cyclic, so it is enough to check the ramified primes. If the discriminant of the polynomial is squarefree then the ramified primes are exactly those primes dividing the discriminant of the polynomial, which in our case is a prime p = 2777. We shall show that G(p) is cyclic by showing that the Galois group of f(x) over Q p, which is isomorphic to G(p), is of order 2. We have the following factorization of f(x) modulo p = 2777: f(x) (x + 787)(x + 929)(x + 1919) 2 (mod 2777) Applying Hensel s Lemma we conclude that f(x) factors into two relatively prime linear factors and one irreducible factor of degree 2 over Q p. Indeed, Hensel s Lemma tells us that f(x) will factor into two relatively prime linear factors and one factor of degree 2, which must be irreducible since f(x) is separable over Q p and p is ramified. This implies that the Galois group of f(x) over Q p is of order 2. We now find two irreducible polynomials g 1 (x), g 2 (x) of degree greater than 1, such that K is the splitting field of their product and the groups U 1 and U 2 are the Galois groups of K over the root fields of g 1 (x) and g 2 (x) respectively. Moreover, the polynomial g 1 (x)g 2 (x) has a root in Q p for all prime p. Since U 1 is the stabilizer of one point, we may take g 1 (x) := f(x). Since U 2 is a 2-sylow subgroup of S 4, we may take g 2 (x) to be the cubic resolvent g 2 (x) = x 3 + 10x 2 + 9x + 1 of f(x). In conclusion, the product: g 1 (x) g 2 (x) = (x 4 5x 2 + x + 4)(x 3 + 10x 2 + 9x + 1) has Galois group S 4 and has a root in Q p for all prime p.

4 DANIEL RABAYEV AND JACK SONN S 5 : The 2-covering of S 5 in [3] is given by U 1 := the stabilizer in S 5 of the set {1,2}, and U 2 := the normalizer N S5 { (1, 2, 3, 4, 5) } of (1, 2, 3, 4, 5) in S 5. f(x) = x 5 9x 3 9x 2 + 3x + 1 has prime discriminant p = 36497. Factoring f(x) mod 3, mod 17, and mod 103 yields the Galois group S 5. The factorization of f(x) modulo p is f(x) (x + 8522)(x + 18488)(x + 19525)(x + 31478) 2 (mod 36497) so (as in the preceding case) the decomposition group at p is cyclic. We now find the desired polynomials g 1 (x), g 2 (x). We start with the polynomial g 1 (x) corresponding to the group U 1, which is the stabilizer in S 5 of the set {1,2}. It is easy to see that U 1 = C 2 S 3, and furthermore, all of the subgroups of S 5 which are isomorphic to C 2 S 3 are conjugates. Hence we need only find a polynomial whose roots lie in K and with Galois group U 1 over its root field. That polynomial is the polynomial whose roots are the sums of pairs of roots of f(x), namely (using GAP), g 1 (x) = x 10 27x 8 9x 7 + 234x 6 + 151x 5 756x 4 585x 3 + 873x 2 + 660x 163 Indeed, it is an irreducible polynomial of degree 10 (= [S 5 : U 1 ]) and according to the previous discussion, it is clear that it has Galois group C 2 S 3 over its root field. We turn next to the polynomial g 2 (x) corresponding to the group U 2 = N S5 (1, 2, 3, 4, 5). The order of U 2 is 20, so g 2 (x) will be of degree 6. First, let us notice that up to conjugation, there is only one subgroup of order 20 in S 5, since every subgroup of order 20 is the normalizer of a 5-cycle. Thus we need only find an irreducible polynomial of degree 6 whose splitting field is contained in K. Our method is ad hoc. Consider the action of S 5 on the set A of the ten 2-subsets of {1, 2, 3, 4, 5}, which is equivalent to its action on the roots of g 1 (x). Then consider the action of S 5 on the set B of 252 5-subsets of A. This latter action has an orbit of length 12 (MAGMA). We wish to produce a separable polynomial of degree 252 with splitting field K such that the action of G(K/Q) = S 5 on its roots is equivalent to its action on B. We start with g 1 (x) = x 10 27x 8 9x 7 + 234x 6 + 151x 5 756x 4 585x 3 + 873x 2 + 660x 163 as above which is an irreducible polynomial with Galois group S 5 over Q. Using a Tschirnausen algorithm 2 in GAP, we obtain a degree 10 polynomial g(x) the sum of whose roots is zero 2 We are grateful to Alexander Hulpke for introducing us to this algorithm in GAP.

2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS 5 and with the same root field (up to conjugacy) as g 1 (x) and such that the polynomial h(x) of degree 252 whose roots are the sums of quintuples of distinct roots of g(x), is separable. We now observe that since the sum of the roots of g(x) is zero, the negative of any root of h(x) is also a root of h(x), so h(x) = r(x 2 ) for some r(t) Q[x] of degree 126. r(t) has an irreducible factor of degree 6 with splitting field contained in K, which is g 2 (x) = x 6 1389834 26569 x5 + 804440497905 705911761 x4 248200341684305820 x 3 18755369578009 + 43053809146811239035780 498311414318121121 x 2 3981047272192561628814509657 x 13239635967018160063849 + 153302536976991922308475759105476 351763888007705494736404081 and is therefore a polynomial of the desired type. S 6 : The 2-covering of S 6 in [3] is given by U 1 := the stabilizer in S 6 of the partition {1,2,3},{4,5,6} and U 2 = S 5. Note that U 1 is of order 72 and is also the normalizer of a 3-Sylow subgroup of S 6. The polynomial f(x) = x 6 10x 4 9x 3 + 5x 2 + 3x 1 has prime discriminant p = 33994921, and factoring mod 13, mod 37, and mod 263 yields the Galois group S 6. f(x) (x + 665896)(x + 3641312)(x + 15713959)(x + 25142575)(x + 11413050) 2 (mod p) so the decomposition group at p is cyclic (as in the preceding case for S 5 ). We will now find the polynomials g 1 (x) and g 2 (x). Let us notice that we already have a polynomial for one of the conjugates of the group U 2, namely g 2 (x) := f(x) = x 6 10x 4 9x 3 + 5x 2 + 3x 1, since the Galois group of f(x) over Q(θ), where θ is a root of f(x), is S 5. We now turn to the polynomial g 1 (x) which corresponds to the group U 1, which is the stabilizer of the partition {1, 2, 3}, {4, 5, 6}. Applying GAP we construct a polynomial h(x) whose roots are the sums of triples of distinct roots of f(x): h(x) = x 20 60x 18 + 1470x 16 19209x 14 + 146153x 12 662097x 10 Observe that h(x) = g 1 (x 2 ), where +1751524x 8 2532942x 6 + 1684385x 4 278413x 2 + 4096. g 1 (x) = x 10 60x 9 + 1470x 8 19209x 7 + 146153x 6 662097x 5 +1751524x 4 2532942x 3 + 1684385x 2 278413x + 4096. g 1 (x) is irreducible over Q. It is evident that the splitting field of h(x) is contained in K, and since g 1 (x) is irreducible over Q, the splitting field of h(x) equals K. It is evident that

6 DANIEL RABAYEV AND JACK SONN h(x) has Galois group S 3 S 3 over Q(η), where η is a root of h(x). The root field Q(η 2 ) of g 1 (x) is a subfield of Q(η) of degree 10 over Q, so Γ := G(K/Q(η 2 )) is a subgroup of order 72 of S 6, containing S 3 S 3. Γ contains a 3-Sylow subgroup of S 6 whose normalizer is of index 1 or 2 in Γ. It now follows from the Sylow theorems the index must be 1, so Γ must be conjugate to U 1 in S 6. (In fact, every subgroup of S 6 of order 72 is conjugate to U 1.) 3. The alternating groups A 4 : A 2-covering of A 4 is given by U 1 := the stabilizer in A 4 of one point, namely, U 1 = C 3, and the group U 2 :=the 2-sylow subgroup of A 4 which is C 2 C 2. It is evident that this is indeed a 2-covering of A 4. Let f(x) = x 4 10x 3 7x 2 + 3x + 2. The discriminant of f(x) is 163 2, a square (of a prime), hence the Galois group is contained in A 4. Factoring mod 3 and mod 5 gives the Galois group A 4. We check that the Galois group of f(x) over Q p is of order 3. The factorization of f(x) factors mod 163 is: f(x) (x + 50)(x + 143) 3 (mod 163) By Hensel s Lemma, f(x) factors into a linear factor and a factor of degree 3 over Q p, which we claim must be irreducible over Q p. Indeed, f(x) is separable over Q p so this cubic factor is either irreducible as claimed or a product of a linear factor and an irreducible factor of degree 2. In the latter case, f(x) has two relatively prime linear factors which implies that the local Galois group fixes exactly two roots, thus contains a transposition which contradicts the fact that this Galois group is contained in A 4. This proves the claim and implies that the Galois group of f(x) over Q p is either S 3 or C 3. The decomposition group cannot be S 3 since it is a subgroup of A 4 and A 4 does not contain a copy of S 3, hence we deduce that it is (the cyclic group) C 3. For the polynomials g 1 (x), g 2 (x), we argue as in the case of S 4. Since U 1 is the stabilizer of one point, we may take g 1 (x) := f(x). Since U 2 is a 2-sylow subgroup of A 4, we may take g 2 (x) to be the cubic resolvent g 2 (x) = x 3 + 89x 2 + 2586x + 24649 of f(x). A 5 : A 2-covering of A 5 can be obtained from a 2-covering of S 5 and the following easy lemma.

2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS 7 Lemma 3.1. [4]. Let G be a 2-coverable group which is covered by the conjugates of the subgroups H and K. If N G and G = NH = NK, then N admits the covering H N, K N. Applying the lemma to the 2-covering of S 5 which was given in a previous section, we take U 1 to be the intersection of A 5 with the stabilizer Stab S5 {1, 2} of the set {1,2} in S 5, and U 2 = N S5 { (1, 2, 3, 4, 5) } A 5. As S 5 = N S5 { (1, 2, 3, 4, 5) } A 5 = Stab S5 {1, 2} A 5, {U 1, U 2 } is a 2-covering of A 6. f(x) = x 5 5x 4 7x 3 + 18x 2 x 1 with discriminant 15733 2, the square of a prime. Factoring mod 3, mod 5, and mod 7 yields the Galois group A 5, and factoring mod the discriminant f(x) (x + 13759)(x + 4886)(x + 4272) 3 (mod 15733), an argument similar to the case of A 4 yields cyclic decomposition groups. We turn now to the polynomials g 1 (x), g 2 (x). From here the proof is similar to the case of S 5. The indices of U 1 and U 2 in A 5 are 10 and 6 respectively. Furthermore, each U i is the unique subgroup of order U i in A 5 up to conjugacy, i = 1, 2. By the same method as in the case of S 5, starting with f(x), we compute and g 1 (x) = x 10 20x 9 + 129x 8 167x 7 1160x 6 + 3139x 5 +2214x 4 9559x 3 + 1089x 2 + 5486x 461 g 2 (x) = x 6 42323896 212521 x5 + 733217461235730 45165175441 + 33179389816151522298285969465 2039893072616309544481 + 91715516822801665361140634800285822768801 x 4 6645087421265775407450 x 3 9598548249896761 x 2 86429398882716670964952901401177395 x 433520115685490720702646601 92132128505596173454447158291121 A 6 : In contrast to the previous examples, in this example not all the decomposition groups turn out to be cyclic. The 2-covering of A 6 is given by applying the lemma from the previous section to the 2-covering of S 6 which was given earlier. Namely, U 1 = H 1 A 6, where H 1 denotes the stabilizer in S 6 of the partition {1, 2, 3}, {4, 5, 6} and U 2 = H 2 A 6 = A 5 where H 2 = S 5. (S 6 = A 6 H 1 = A 6 H 2 because U 1 and U 2 each contains an odd permutation.). f(x) = x 6 3x 5 4x 4 + 5x 3 + 3x 2 5x 1,

8 DANIEL RABAYEV AND JACK SONN with discriminant 14341 2, the square of a prime. Factoring mod 3, mod 5, and mod 101 yields the Galois group A 6. In order to verify that the requirements hold, we need the decomposition group of a prime above p = 14341 to be contained in a conjugate of U 1 or U 2. We claim that the Galois group of f(x) over Q p is contained in a conjugate of U 2. f(x) factors mod p as a product of two distinct linear factors and the square of an irreducible quadratic factor: f(x) (x + 5464)(x + 5605)(x 2 + 8805x + 9098) 2 (mod 14341) By Hensel s Lemma, f(x) factors into two relatively prime linear factors and one irreducible factor of degree 4 over Q p. This means that the decomposition group, which is contained in A 6, fixes two roots of f(x), and is therefore contained in a conjugate of A 5 = U 2. Note that the decomposition group is noncyclic, since its order is divisible by 4 and is embedded in A 4. We turn to the polynomials g 1 (x) and g 2 (x). First, as U 2 = A 5, we may take g 2 (x) = f(x) = x 6 3x 5 4x 4 + 5x 3 + 3x 2 5x 1. By the same method used for S 6, we arrive at the polynomial g 1 (x) = x 10 93 2 x9 + 14253 1886832143 2048 16 x8 75007 8 x 3 + 45377978093 65536 x 7 + 7698057 128 x 6 62688663 256 x 2 33044483389 x + 1696039489 131072 1048576 x 5 + 1276887561 x 4 2048 A 7 : The 2-covering of A 7 given in [4] is U 1 = N A7 (1234567), of order 21, and U 2 = (Sym{1, 2} Sym{3, 4, 5, 6, 7}) A 7, of order 120, where Sym{1, 2} is the symmetric group on {1, 2} and Sym{3, 4, 5, 6, 7} is the symmetric group on {3, 4, 5, 6, 7}. f(x) = x 7 2x 6 5x 5 x 4 3x 3 x 2 x 5 with discriminant 554293 2, the square of a prime. The factorizations of f(x) modulo 3, 5, 7, 23 show that the Galois group contains permutations of the types [7],[5],[2,4],[3,3] respectively. As A 7 has no proper subgroup of order divisible by 3 4 5 7 (GAP), the Galois group is A 7. We show next that the Galois group of f(x) over Q p is contained in a conjugate of U 2, where p = 554293 is the unique ramified prime. The factorization of f(x) modulo p is f(x) (x + 521396)(x + 134869) 2 (x + 281696) 2 (x 2 + 308251x + 256808) (mod 554293). Applying Hensel s Lemma we conclude that f(x) factors into a (separable) product of a linear factor and quadratic factors over Q p. It follows that the decomposition group G(p) is a 2-group. But this means that the decomposition group is contained in a conjugate of U 2 since U 2 contains a 2-Sylow subgroup of A 7. For the explicit polynomials g 1 (x), g 2 (x), we start with g 1 (x) and observe first that U 1 is not a maximal subgroup of A 7. It is contained in a maximal subgroup of order 168, isomorphic to GL 3 (2). Although the unique conjugacy class of subgroups of order 168 in S 7 splits into

2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS 9 two conjugacy classes in A 7, each contains the unique conjugacy class of subgroups of order 21 in A 7. It follows that either of the two conjugacy classes of subgroups of order 168 in A 7 can replace the conjugacy class of U 1 in the given 2-covering of A 7. It is advantageous to do this in order to seek a g 1 (x) of smaller degree (15) than that of the g 1 (x) (120) that would correspond to the original U 1. The method for seeking the explicit g 1 (x) and g 2 (x) is similar to the preceding, except for a greater dependence on trial and error, and that in addition to the two algorithms in GAP used earlier, the Tschirnhausen algorithm and the algorithm that produces a polynomial whose roots are sums of roots of a given polynomial, an algorithm that produces a polynomial whose roots are products of roots of a given polynomial was pressed into service. The polynomials corresponding to the groups U 1 and U 2 are: g 1 (x) = x 15 + 236441 835 x14 553005 3796 x13 142958 24613 x12 + 3151297 45345 x11 5686331 48267 x10 + 30589 7924 x9 + 10004127 37216 x8 255049 586 x7 41630027 36825 x6 + 72184440 x 5 8779 45952163 35584 x4 1964679 5893 x3 + 14351259 x 2 3135356x 791507 5423 2567 457 respectively. g 2 (x) = x 21 30x 20 + 395x 19 2937x 18 + 12917x 17 29077x 16 6543x 15 +236441x 14 553005x 13 142958x 12 + 3151297x 11 5686331x 10 +305889x 9 + 10004127x 8 255049x 7 41630022x 6 + 72184440x 5 45952163x 4 1964679x 3 + 14351259x 2 3135356x 791507 A 8 : The 2-covering of A 8 in [4] is given by U 1 :=the affine linear group on F 3 2 acting as permutations on the 8 points of this space (shown to be embedded into A 8 ), isomorphic to F 3 2 GL 3 (F 2 ), of index 15, and U 2 := [Sym{1, 2, 3} Sym{4, 5, 6, 7, 8}] A 8, of index 56. f(x) = x 8 x 7 2x 6 x 5 + 3x 4 + 3x 3 + 2x 2 + x + 1 with discriminant 11489 2, the square of a prime. The factorizations of f(x) modulo 3, 37, 41 show that the Galois group contains permutations of the types [7], [4, 4], [3, 5] respectively. As A 8 has no proper subgroup of order divisible by 3 4 5 7 (GAP), the Galois group is A 8. Now f(x) factors modulo p as: f(x) (x + 1440) 3 (x 2 + 10435x + 8884)(x 3 + 8222x 2 + 9218x + 10584) (mod 11489) Hensel s Lemma yields a corresponding factorization f(x) = a(x)b(x)c(x) over Q p with a(x), c(x) of degree 3 and b(x) of degree 2. The splitting field K p of f(x) over Q p is the composite of the splitting fields K a, K b, K c of these three factors, the first of which is ramified and the others unramified. a(x) cannot be a product of linear factors since p ramifies in the splitting field of f(x). Thus a(x) is either an irreducible cubic or a product of a linear and a quadratic polynomial. b(x) and c(x) are irreducible. If a(x) is a product of a linear and a quadratic polynomial, then K a /Q p is a ramified quadratic extension and

10 DANIEL RABAYEV AND JACK SONN G(K p /Q p ) = C 2 C 2 C 3, and contains transpositions, contrary to the fact that it is contained in A 8. Therefore a(x) is an irreducible cubic. It follows that G(K a /Q p ) is C 3 or S 3. C 3 is not possible because K a /Q p would then be totally and tamely ramified, but p 2 (mod 3) hence Q p does not contain the cube roots of unity. Hence G(K a /Q p ) is S 3. The only way that G(K p /Q p ) can be contained in A 8 is if K b is contained in K a as the unramified quadratic subfield, and the action of G(K a /Q p ) on the roots of a(x)b(x) is the representation of S 3 as (123), (12)(45). Furthermore, since K c /Q p is unramified of degree 3, K a K c = Q p, so G(K p /Q p ) acts on the roots of f(x) as (123), (12)(45), (678), which is contained in U 2. A trial and error search similar to that in the preceding section yields the following polynomials corresponding to the groups U 1 and U 2 : and g 1 (x) = x 15 95 2 x14 + 3197 4 x13 210405 x 12 7433761 x 11 + 22263473 x 10 104363621 x 9 32 256 256 512 11710437647 x 8 + 918294939277 x 7 + 5512444210853 x 6 21829527485865 x 5 410948046145887 x 4 4096 65536 131072 262144 2097152 14737903234907309 x 3 27794916917226379 x 2 31865126119189869 x 1256672780150364253 16777216 8388608 4194304 134217728 g 2 (x) = x 56 21x 55 + 180x 54 735x 53 + 783x 52 + 5407x 51 21690x 50 + 6260x 49 + 134354x 48 256270x 47 304036x 46 + 1275397x 45 + 942277x 44 6877457x 43 2069019x 42 + 42340751x 41 37776712x 40 158592858x 39 + 379188834x 38 + 107601914x 37 1455434817x 36 + 1392744196x 35 +2854906257x 34 7385711268x 33 + 817821146x 32 + 17194579246x 31 19188201050x 30 18578454420x 29 + 56645526134x 28 18919098068x 27 77381156815x 26 + 94740276783x 25 +30587068228x 24 150642640537x 23 + 94400959739x 22 + 77653585985x 21 155812844138x 20 +65943540240x 19 + 68936866008x 18 121634333784x 17 + 85946585062x 16 14385996295x 15 52121609885x 14 + 76576460155x 13 37676127233x 12 34141244957x 11 + 82692304838x 10 respectively. 86059996214x 9 + 64840248670x 8 43929749632x 7 + 29390580877x 6 17890022300x 5 +8764351999x 4 3189669694x 3 + 811672328x 2 129075408x + 9546832 References [1] D. Berend and Y. Bilu, Polynomials with roots modulo every integer, Proc. AMS 124 (1996) 1663 1671 [2] R. Brandl, Integer polynomials with roots mod p for all prime p, J. Alg 240 (2001) 822 835 [3] R. Brandl, D. Bubboloni and I. Hupp, Polynomials with roots mod p for all prime p, J. Group Th. 4 (2001) 233 239 [4] D. Bubboloni, Coverings of the symmetric and alternating groups, preprint, Quaderno del Dipartimento di Matematica U. Dini, Firenze 7 (1998). [5] J. Sonn, SL(2, 5) and Frobenius Galois groups over Q, Can. J. Math. 32 (1980) 281 293 [6] J. Sonn, Polynomials with roots in Q p for all p, Proc. AMS 136 (2008) 1955 1960

2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS 11 Department of Mathematics, Technion Israel Institute of Technology, Haifa, 32000, Israel E-mail address: danielr@tx.technion.ac.il Department of Mathematics, Technion Israel Institute of Technology, Haifa, 32000, Israel E-mail address: sonn@tx.technion.ac.il